Veritas Quant Question

[fermar84]

Hi, could you please explain this one? Thank you
If a and b are integers and (ab)^5=96y,y could be

(a) 5
(b) 9
(c) 27
(d) 81
(e) 125

[Test Prep Unlimited]

The prime factorization of 96 is 2^5*3. So since a and b are integers, (ab)^5 must have at least 2^5 and 3^5 as factors. Well, 96 has 2^5 and one 3, so 3^4 is left over (which is y). 3^4=81.

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[j_shreyans]

Hi ,

(ab)^5=96y

We can write 96=2^5X3

So

a^5b^5=2^5 x 3xy

We need four more 3 to equate the equation.

So y=3^4 which is 81

OAD

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Hi Fernando,

This question is essentially about prime-factorization - the idea that any positive integer greater than 1 is either prime or the product of a bunch of primes.

Here, we're told that (AB)^5 = 96Y, and that A and B are integers, which means...

(AB)(AB)(AB)(AB)(AB) = 96Y

We can rewrite this as....

(A^5)(B^5) = (2^5)(3)(Y)

We're asked for what Y COULD equal. This means that Y could be MORE than one value...we should start by looking for the smallest value that Y could equal.

Notice how 2^5 could "account for" either A or B, so we need to make sure that the "Y", when combined with the "3" that's already there, could account for the other variable....

If Y = 3^4, then 96Y would = (2^5)(3^5), which gives us two integers raised to the 5th power.

Y COULD = 3^4 = 81

Final Answer: D

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